SOLUTION: Find all the real solutions of the equation. x^5+8x^2=0

Algebra ->  Equations -> SOLUTION: Find all the real solutions of the equation. x^5+8x^2=0      Log On


   



Question 706539: Find all the real solutions of the equation.
x^5+8x^2=0

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E5%2B8x%5E2=0
Solving equations like these is done by getting one side of the equation equal to zero, then factoring the other side and then using the Zero Product Property to get solvable equations.

Since we already have one side equal to zero we can proceed to the factoring. When factoring always start with the greatest common factor (GCF). (Note: If the GCF is 1 then it is usually not factored out.) The GCF here is x%5E2:
x%5E2%28x%5E3%2B8%29+=+0

After the GCF then you try any or all of the other factoring techniques you have learned. One of these techniques is to use factoring patterns. And one of these patterns is for a sum of cubes: a%5E3%2Bb%5E3+=+%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29. Since x%5E3 is an obvious cube and since 8+=+2%5E3 our second factor is a sum of the cubes of x and 2. Using "x" for the "a" and "2" for the "b" in the pattern we get:
x%5E2%28x%2B2%29%28x%5E2%2Bx%2A2%2B2%5E2%29+=+0
which simplifies to:
x%5E2%28x%2B2%29%28x%5E2%2B2x%2B4%29+=+0
None of the factoring techniques will factor the third factor so we are finished factoring.

The Zero Product Property tells us that a product that equals zero (like we have) can only happen if one of the factors is zero. So:
x%5E2+=+0 or x%2B2=0 or x%5E2%2B2x%2B4+=+0
The first two equations are easy to solve: x = 0 or x = -2. For the third one we will need the Quadratic Formula:
x+=+%28-%282%29+%2B-+sqrt%28%282%29%5E2+-+4%281%29%284%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%282%29+%2B-+sqrt%284+-+4%281%29%284%29%29%29%2F2%281%29
x+=+%28-%282%29+%2B-+sqrt%284+-+16%29%29%2F2%281%29
x+=+%28-%282%29+%2B-+sqrt%28-12%29%29%2F2%281%29
At this point we can stop. With the -12 in the square root, the solutions to this third equation will be complex, not real. (You can't square any real number and get -12!) The problem specifically asks for only real solutions so we stop trying to solve this third equation.

So our only real solutions come from the other two equations:
x = 0 or x = -2